Difference of Two Squares

Spot a square minus a square and factorise it into two brackets, one with minus and one with plus.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel IGCSE Maths 4MA1, IGCSE Additional Maths 0606 · 24 real past-paper questions below

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What a Difference of Two Squares Is

A difference means a take-away. So a difference of two squares is one square number or term minus another, like $x^{2} - 49$ .

Formula

a^{2} - b^{2} = (a - b) (a + b)

Here $a$ is what is squared in the first term and $b$ is what is squared in the second. One bracket gets a minus, the other a plus.

Worked example

1
$(x - 7) (x + 7)$ — Start with two brackets that have the same 7 but opposite signs. Let us see why they give a difference of two squares.
2
$= x^{2} + 7 x - 7 x - 49$ — Multiply every term in the first bracket by every term in the second: $x \times x$ , $x \times 7$ , $- 7 \times x$ , $- 7 \times 7$ .
3
$= x^{2} + 7 x - 7 x - 49$ — The two middle terms are the same size with opposite signs, so they add to 0 and cancel.
4
$= x^{2} - 49$ — Only the two squares are left: $x^{2}$ minus $7^{2} = 49$ .

Key idea

Factorising just runs this backwards: see $x^{2} - 49$ , write $(x - 7) (x + 7)$ .

Key idea

Work out $101 \times 99$ in your head. Write it as $(100 + 1) (100 - 1)$ , which is $10 0^{2} - 1^{2} = 10000 - 1 = 9999$ . Try $52 \times 48$ the same way.

Square Minus Square

The 2-mark question: factorise something like $25 x^{2} - 9 y^{2}$ . Ask what is squared in each term.

Worked example

1
$25 x^{2} - 9 y^{2}$ — The question. Two terms, a minus between them, and both are squares. That is the sign to use this pattern.
2
$= (5 x)^{2} - (3 y)^{2}$ — Square root each term: $25 x^{2}$ is 5x and $9 y^{2}$ is 3y. Root the number and halve the power.
3
$= (5 x - 3 y) (5 x + 3 y)$ — Now use $a^{2} - b^{2} = (a - b) (a + b)$ with a = 5x and b = 3y. The order of the brackets does not matter.

Common mistake

Writing $(25 x - 9 y) (25 x + 9 y)$ . Use the square roots: $25 x^{2} = 5 x$ , not $25 x$ .

Take Out a Number First

When the question says factorise fully, the numbers may not be square yet. Take out the common factor first, then look again.

Worked example

1
$2 x^{2} - 72$ — The question. 72 is not a square number, so the pattern will not work yet.
2
$= 2 (x^{2} - 36)$ — Both 2 and 72 divide by 2, so take it out: $2 x^{2} \div 2 = x^{2}$ and $72 \div 2 = 36$ .
3
$= 2 (x - 6) (x + 6)$ — Now the bracket is a square minus a square: $36 = 6^{2}$ . The 2 stays in front.

Exams often use two letters and hide an x in the common factor too. Same idea.

Worked example

1
$3 x^{3} - 48 x y^{2}$ — The question. 48 is not square, and $x^{3}$ is not a square either.
2
$= 3 x (x^{2} - 16 y^{2})$ — Both terms divide by 3 and by x, so take out 3x: $3 x^{3} \div 3 x = x^{2}$ and $48 x y^{2} \div 3 x = 16 y^{2}$ .
3
$= 3 x (x - 4 y) (x + 4 y)$ — Inside is a difference of two squares: $16 y^{2}$ is 4y squared. One minus bracket, one plus bracket.

Common mistake

Stopping at $2 (x^{2} - 36)$ , or writing $(2 x - 12) (x + 6)$ . Both are right but not finished: the first bracket still factorises, the second still has a 2 inside it.

Key idea

Show the common-factor line. If you make a slip after it, that line alone can still earn you a mark.

Use It on Numbers

Some questions ask for $5 3^{2} - 4 7^{2}$ without a calculator. Squaring both is slow. Use the pattern.

Worked example

1
$5 3^{2} - 4 7^{2}$ — The question. It is a square minus a square, so $a^{2} - b^{2} = (a - b) (a + b)$ works on numbers too.
2
$= (53 - 47) (53 + 47)$ — Put a = 53 and b = 47 into the two brackets: one minus, one plus.
3
$= 6 \times 100$ — Work out each bracket: 53 − 47 = 6 is the difference and 53 + 47 = 100 is the sum.
4
$= 600$ — Multiplying by 100 is easy, which is why this is faster than squaring 53 and 47.

Key idea

Look for a sum that makes a round number, like 100. That is the sign the question wants this trick.

Twice, or a Bracket Squared

Harder questions hide the squares. A power of 4 is a square of a square, and a whole bracket can be the thing that is squared.

Worked example

1
$x^{4} - 81 = (x^{2})^{2} - 9^{2}$ — Write each term as a square: $x^{4}$ is x² squared, because $x^{2} \times x^{2} = x^{4}$ , and 81 is 9 squared.
2
$= (x^{2} - 9) (x^{2} + 9)$ — Use the pattern once, with a = x² and b = 9.
3
$= (x - 3) (x + 3) (x^{2} + 9)$ — Check each bracket again. $x^{2} - 9$ is also a square minus a square, so it splits into $(x - 3) (x + 3)$ .

Worked example

1
$(x + 5)^{2} - 4^{2}$ — The question. Here the whole bracket x + 5 is the thing being squared, so treat it as a.
2
$= (x + 5 - 4) (x + 5 + 4)$ — Use the pattern with a = x + 5 and b = 4: one minus, one plus.
3
$= (x + 1) (x + 9)$ — Tidy each bracket: 5 − 4 = 1 and 5 + 4 = 9.

After tidying, a bracket may still have a common factor. Fully means you take it out.

Worked example

1
$(3 x + 1)^{2} - 16$ — The question. $16 = 4^{2}$ , and the bracket 3x + 1 is squared.
2
$= (3 x + 1 - 4) (3 x + 1 + 4)$ — Use the pattern with a = 3x + 1 and b = 4.
3
$= (3 x - 3) (3 x + 5)$ — Tidy each bracket: 1 − 4 = −3 and 1 + 4 = 5.
4
$= 3 (x - 1) (3 x + 5)$ — The first bracket 3x − 3 divides by 3, so take the 3 out to the front. 3x + 5 has no common factor, so stop.

Common mistake

Trying to factorise $x^{2} + 9$ . A square plus a square never factorises.

Which Method? Read the Question

Check the sign between the terms and whether each term is a square.

You see	You do
$25 x^{2} - 9 y^{2}$	Square roots, one minus bracket, one plus
“Fully”, numbers not square	Take out the common factor first
$5 3^{2} - 4 7^{2}$	Difference times sum
$x^{4}$ or a bracket squared	Factorise, then check each bracket again
A square plus a square	It does not factorise
A fraction like $\frac{x - 4}{x ^{2} - 16}$	Factorise the bottom, then cancel. See Algebraic Fractions
Three terms, like $x^{2} + 5 x + 6$	See Factorising Quadratic Expressions

Key idea

Check any answer by expanding: the middle terms must cancel.

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